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Sketch graphs of

Casio

Member
Feb 11, 2012
86
I have a bit of a misunderstanding with;


y = (x - 2)^2


I understand it to be a quadratic, and if I used the formula to work it out I would see two roots,


x = - 2 or x = 2


If I put the above equation into a graphics calculator the result is always x = 2


Looking at the equation above I could just say x^2 and 2^2 = x^2 + 4


What I don't understand is by looking at the equation how x = 2 when it is graphed?


I know its right but don't understand?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
To sketch the graph, note that:

y = (x-2)2

is the graph of y = x2 translated two units to the right.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Here is what the graph looks like.

[graph]xzrhkxtmh3[/graph]

We can see it only has one root so can you show us how you solved it algebraically?
 

Casio

Member
Feb 11, 2012
86
To sketch the graph, note that:

y = (x-2)2

is the graph of y = x2 translated two units to the right.
I know you are right mark and the graphs I have are as you say the same, but translated two units to the right, which is a part I don't understand?
 

Casio

Member
Feb 11, 2012
86
Here is what the graph looks like.

[graph]xzrhkxtmh3[/graph]

We can see it only has one root so can you show us how you solved it algebraically?
I am not going to say this is absolutely correct and if not please do put me on the right tracks.

y = (x - 2)2

First I expanded the brackets;

(x - 2)(x - 2)

Then I multiplied them out;

x2- 2x - 2x + 4

then,

x2 - 4x + 4

At this point I thought I have a quadratic

ax2 + bx + c = 0

I used the formula

b + or - square root b2 - 4(a)(c)/2a

I put the data in;

4 + or - square root 42 - 4 x 4 / 2(1)

I ended up with

-4/2 = -2 or 4/2 = 2

Now I am not 100% sure if I am understanding the algebra with regards to;

ax2 + bx + c = 0 and

x2 - 4x + 4

I am not sure when using the above ax2 in relation to x2 whether I have 1 or 0 at that point?

I assumed 1, which is how I ended up with two roots.

But am unclear.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We know y = x2 has its axis of symmetry where x = 0. Then y = (x - h)2 will have its axis of symmetry where:

x - h = 0 or x = h.

This is a very common point of confusion for students of algebra. It seems to go against intuition that f(x - h) moves the graph of f(x) h units to the right, when h is being subtracted from x. What in fact happens is the axes are translated to the left, relative to the graph, which is the same as the graph being translated to the right relative to the axes.

You know that a vertical translation is y = f(x) + k. This moves the graph of y = f(x) up k units. This agrees nicely with intuition, since we are adding to the original to move it up. But, we may also write the translation as y - k = f(x).

Does this make more sense now?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
With regards to finding the roots, we have:

y = (x - 2)2

To find the roots, we set y = 0, and have:

0 = (x - 2)2

Take the square root of both sides, noting that ±0 = 0.

x - 2 = 0

x = 2

Thus, we have the repeated root x = 2.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Casio,

As MarkFL showed this is actually much easier than the way you did it, which is always nice :) Sometimes we can over complicate things in math.

\(\displaystyle (x-2)^2=0\)

Why do we set this equal to 0? Well, the "roots" of a quadratic are the values of x that correspond to y=0. So we simply set y=0 and solve for x. Starting with the above equation we get that:

(1) \(\displaystyle (x-2)^2=0\)
(2) \(\displaystyle x-2 =0\) (take the square-root of both sides, noting that $\sqrt{0}=0$)
(3)\(\displaystyle x=2\)

Thus we get only 1 solution. Again, this is exactly what MarkFL showed.
 

Casio

Member
Feb 11, 2012
86
We know y = x2 has its axis of symmetry where x = 0. Then y = (x - h)2 will have its axis of symmetry where:

x - h = 0 or x = h.

This is a very common point of confusion for students of algebra. It seems to go against intuition that f(x - h) moves the graph of f(x) h units to the right, when h is being subtracted from x. What in fact happens is the axes are translated to the left, relative to the graph, which is the same as the graph being translated to the right relative to the axes.

You know that a vertical translation is y = f(x) + k. This moves the graph of y = f(x) up k units. This agrees nicely with intuition, since we are adding to the original to move it up. But, we may also write the translation as y - k = f(x).

Does this make more sense now?
Thanks Mark, however you have introduced functions I think, which has made the understanding at the moment a little more difficult as I have not yet covered that module.

I think my misunderstanding is in the idea of what I should do to the equation;

y = (x - 2)2

I look at the above and I say everything in the brackets are squared, so;

y = x2 - 22 , and this becomes; x2 - 4

Looking at the above I see a quadratic graph at y = - 4, which I know is incorrect.

So I am looking for a mathematical rule which tells me the correct way that a given branch of mathematics should be worked out.

The administrator did it this way;

y = (x - 2)2
0 = (x - 2)2
sqrt 0 = x - 2

x = 2

so 2 has moved to the opposite side of x, which I have seen many times before but am unsure when to use which order of working out, i.e the long winded way I did previously or this short quick method which I did not know about?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have made a very common error, sometimes referred to as "the freshman's dream," which is to state:

(x - 2)2 = x2 - 22

This is not true, what you actually have is:

(x - 2)2 = x2 - 4x + 4

If you have been taught the FOIL method, try this with:

(x - 2)(x - 2)

and you will find the above result.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I am not going to say this is absolutely correct and if not please do put me on the right tracks.

y = (x - 2)2

First I expanded the brackets;

(x - 2)(x - 2)

Then I multiplied them out;

x2- 2x - 2x + 4

then,

x2 - 4x + 4

At this point I thought I have a quadratic

ax2 + bx + c = 0

I used the formula

b + or - square root b2 - 4(a)(c)/2a

I put the data in;

4 + or - square root 42 - 4 x 4 / 2(1)

I ended up with

-4/2 = -2 or 4/2 = 2
Hi casio, :)

The approach you have taken in finding the roots is correct although a much more easier method is suggested by Jameson. The part where you have made a mistake is highlighted above. Check it again,

\[x=\frac{4\pm \sqrt{4^2 - (4\times 4)}}{2\times 1}\]

Now I am not 100% sure if I am understanding the algebra with regards to;

ax2 + bx + c = 0 and

x2 - 4x + 4

I am not sure when using the above ax2 in relation to x2 whether I have 1 or 0 at that point?

I assumed 1, which is how I ended up with two roots.

But am unclear.
\(ax^2+bx+c=0\) is the general form of a quadratic equation. Compare the quadratic equation \(x^2-4x + 4\) with the general form and see what are the coefficients. Then, \(a=1,\,b=-4\mbox{ and }c=4\). So you are correct in mentioning \(a=1\).

Kind Regards,
Sudharaka.